Question

I have the following function written in Go. The idea is the function has a string passed to it and returns the first IPv4 IP address found. If no IP address is found, an empty string is returned.

func parseIp(checkIpBody string) string {
    reg, err := regexp.Compile("[0-9]+\.[0-9]+\.[0-9]+\.[0-9]+")
    if err == nil {
        return ""
    }   
    return reg.FindString(checkIpBody)
}

The compile-time error I’m getting is

unknown escape sequence: .

How can I tell Go that the '.' is the actual character I’m looking for? I thought escaping it would do the trick, but apparently I’m wrong.

Answer 1

The \ backslash isn’t being interpreted by the regex parser, it’s being interpreted in the string literal. You should escape the backslash again:

regexp.Compile("[0-9]+\\.[0-9]+\\.[0-9]+\\.[0-9]+")

A string quoted with " double-quote characters is known as an "interpreted string literal" in Go. Interpreted string literals are like string literals in most languages: \ backslash characters aren’t included literally, they’re used to give special meaning to the next character. The source must include \\ two backslashes in a row to obtain an a single backslash character in the parsed value.

Go has another alternative which can be useful when writing string literals for regular expressions: a "raw string literal" is quoted by ` backtick characters. There are no special characters in a raw string literal, so as long as your pattern doesn’t include a backtick you can use this syntax without escaping anything:

regexp.Compile(`[0-9]+\.[0-9]+\.[0-9]+\.[0-9]+`)

These are described in the "String literals" section of the Go spec.